parmenides51 wrote: Determine all functions $f$ on the set of positive rational numbers such that $f(xf(x) + f(y)) = f(x)^2 + y$ for all positive rational numbers $x, y$. Let $P(x,y)$ be the assertion $f(xf(x)+f(y))=f(x)^2+y$ Let $a=f(1)^2$ $f(x)$ is injective. 1) $f(x+y)=f'x)+h(y)$ for some $h(x)$ from $\mathbb Q^+\to\mathbb Q$ $P(1,x)$ $\implies$ $f(f(1)+f(x))=x+a$ Let $t\in\mathbb Q^+$ Let $u(t),v(t)\in\mathbb Q^+$ such that : $u(t)+v(t)=\left\lceil\sqrt t\right\rceil$ $u(t)-v(t)=\frac{t}{\left\lceil\sqrt t\right\rceil+2a}$ Let $u_1(t)=f(1)+f(u(t))$ and $v_1(t)=f(1)+f(v(t))$ Let $u_2(t)=u_1(t)f(u_1(t))$ and $v_2(t)=v_1(t)f(v_1(t))$ $t=(u(t)-v(t))(u(t)+v(t)+2a)$ $=(u(t)+a)^2-(v(t)+a)^2$ And so $t=f(f(1)+f(u(t)))^2-f(f(1)+f(v(t)))^2$ $=f(u_1(t))^2-f(v_1(t))^2$ And so $x+t+f(v_1(t))^2=x+f(u_1(t))^2$ $P(u_1(t),x)$ $\implies$ $f(u_2(t)+f(x))=f(u_1(t))^2+x$ $P(v_1(t),x+t)$ $\implies$ $f(v_2(t)+f(x+t))=f(v_1(t))^2+x+t$ And so $f(u_2(t)+f(x))=f(v_2(t)+f(x+t))$ And so (injectivity $u_2(t)+f(x)=v_2(t)+f(x+t)$ Which is $f(x+t)=f(x)+u_2(t)-v_2(t)$ Q.E.D. 2) $f\equiv Id$ Let $Q(x,y)$ be the assertion $f(x+y)=f(x)+h(y)$ Subtracting $Q(x,y)$ from $Q(y,x)$, we get $h(x)-f(x)=h(y)-f(y)$ $=c$ constant. So $h(x)=f(x)+c$ And $Q(x,y)$ becomes $f(x+y)=f(x)+f(y)+c$ And $f(x)+c$ is additive and so ($\mathbb Q$) linear Plugging $f(x)=dx-c$ in original equation, we get $c=0$ and $d=1$ (the case $d=-1$ is non consistant with the fact that $f(x)$ is a positive rational. And so $\boxed{f(x)=x\quad\forall x}$ which indeed is a solution

SPLENDID SOLUTION!